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I've been studying functional analysis and currently solving problems in "Mathematcial Methods in Quantum Mechanics With Applications to Schrodinger Operators" written by G. Teschl, but I have a difficulty in proving the following problem(Problem 3.7 in the book).

Show that for a self-adjoint operator $A$ we have $\Vert AR_A(z)\Vert \le \frac{|z|}{|Im(z)|}, $ where $R_A(z) = (A-z)^{-1}$ is the resolvent.

It seems radical to me. I do not know where to start. I tried $AR_A(z) = (z+A-z)R_A(z) = zR_A(z) + I$, but it did not proceed further. I guess the resolvent formula $R_A(z) - R_A(z') = (z-z')R_A(z)R_A(z')$ will play the significant role in solving the problem.

Can someone give me a help? Any idea or hint would be aprreciated a lot.

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I cannot think of a pretty way to do it (fairly sure there has to be). But, you now that $AR_A(z)$ is normal. So $$ \|AR_A(z)\|=\max\{|\lambda|:\ \lambda\in \sigma(AR_A(z))\}=\max\left\{\frac{|\lambda|}{|\lambda-z|}:\ \lambda\in\sigma(A)\right\}. $$ Since $A$ is selfadjoint, we know that $\sigma(A)\subset\mathbb R$. Write $z=a+ib$. For your formula to work we need $b\ne0$. Thus we want to maximize $$ \frac{|t|}{|t-a+ib|}=\frac{|t|}{\sqrt{(t-a)^2+b^2}}. $$ Consider the square, we want to maximize the function $$ t\longmapsto \frac{t^2}{(t-a)^2+b^2}. $$ Differentiating and equating to zero gives you that the above function has critical points at $t=0$ and $t=(a^2+b^2)/a$. So $$ \|AR_A(z)\|\leq\sqrt{\frac{(a^2+b^2)^2/a^2}{((a^2+b^2)/a-a)^2+b^2}}=\sqrt{\frac{a^2+b^2}{b^2}}=\frac{|z|}{|\operatorname{Im} z|} $$

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  • $\begingroup$ This is insane. Surprised by your insight and computation. Thanks Argerami. $\endgroup$ – Euduardo Dec 9 '18 at 5:43

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